What is the extraneous solution to these equations? $\dfrac{x^2 + 11}{x - 7} = \dfrac{15x - 45}{x - 7}$
Solution: Multiply both sides by $x - 7$ $ \dfrac{x^2 + 11}{x - 7} (x - 7) = \dfrac{15x - 45}{x - 7} (x - 7)$ $ x^2 + 11 = 15x - 45$ Subtract $15x - 45$ from both sides: $ x^2 + 11 - (15x - 45) = 15x - 45 - (15x - 45)$ $ x^2 + 11 - 15x + 45 = 0$ $ x^2 + 56 - 15x = 0$ Factor the expression: $ (x - 7)(x - 8) = 0$ Therefore $x = 7$ or $x = 8$ At $x = 7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 7$, it is an extraneous solution.